= Solution
First pass from the <preorder> $R$ to a <partial order>. Define the <indifference relation> $y\sim z$ by $yRz$ and $zRy$. <Reflexivity> and <transitivity> make this an <equivalence relation>. On the quotient set $C=\mathbb R^L/{\sim}$, put
$$
[y]\unrhd[z]\quad\Longleftrightarrow\quad yRz.
$$
<Transitivity> makes this independent of the representatives, and the quotient relation is antisymmetric, hence a <partial order>.
We need its order-extension property, including freedom to order incomparable classes. Here is the argument. Order all <partial order> extensions of $\unrhd$ by inclusion. The union of a chain of extensions is again a <partial order>: any finite list of comparisons needed to check an axiom occurs together in one member of the chain. <Zorn's lemma> therefore supplies a maximal extension. If two classes $A,B$ remain incomparable, adjoin $A\unrhd B$ and take the <transitive closure of a relation>. The only possible new comparisons have the form $X\unrhd A\unrhd B\unrhd Y$. A cycle would require a previously existing path $B\unrhd A$, contradicting incomparability. The enlarged relation is still a <partial order>, contradicting maximality. Consequently the maximal extension is a <total order>. This also proves the <Szpilrajn extension theorem> used here.
Lift such a <total order> to the original bundles by comparing their equivalence classes. The lifted <preference relation> $R'$ is complete and <transitive>, contains $R$, and preserves its <strict preference>: if $yPz$, their distinct classes satisfy $[y]\unrhd[z]$, and antisymmetry of the extended order prevents the reverse comparison. In particular, the class $\mathcal P$ is nonempty and $R\subseteq\bigcap_{R'\in\mathcal P}R'$.
For the reverse inclusion, take a pair with $\neg(yRz)$. If $zRy$, then $zPy$, so every member of $\mathcal P$ strictly ranks $z$ above $y$ and excludes $yR'z$. If neither comparison holds, their classes are incomparable. First adjoin $[z]\unrhd[y]$, which creates no cycle by the preceding argument, and then extend to a <total order>. Its lift belongs to $\mathcal P$ and again excludes $yR'z$. Thus every pair outside $R$ is outside at least one member of $\mathcal P$, proving
$$
\boxed{\bigcap_{R'\in\mathcal P}R'=R.}
$$
The intersection proof is an order-theoretic fact about any <preorder>. The <congruence axiom> additionally ensures exact rationalization of the observed <demand correspondence>: every chosen bundle weakly outranks every affordable bundle, while an affordable unchosen bundle is strictly worse by part (b). Hence each completion's budget-maximizing set is exactly the observed chosen set.
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